Solution Explanation: Maximum Earnings From Taxi

This problem asks to find the maximum earnings from taxi rides, given a set of rides with start points, end points, and tips. The key constraint is that you can only drive one passenger at a time, and you must proceed in the direction from point 1 to point n.

We can solve this problem efficiently using dynamic programming or memoization with binary search. Both approaches have a time complexity of O(m log m), where m is the number of rides. The space complexity is O(m) for both.

Approach 1: Memoization with Binary Search

1. Sort Rides: First, sort the rides array by start point (start_i). This helps in efficiently finding the next available ride after completing a current ride.

2. Recursive DFS (with Memoization): A recursive function dfs(i) calculates the maximum earnings starting from the i-th ride.

   • Base Case: If i is beyond the number of rides, the earnings are 0.
   • Recursive Step: For each ride i, we have two choices:
     • Skip the ride: The maximum earnings are dfs(i + 1).
     • Take the ride: The earnings are end_i - start_i + tip_i. To find the next available ride, we perform a binary search on the sorted rides array to find the index j of the first ride whose start_j is greater than or equal to end_i. The total earnings in this case become dfs(j) + end_i - start_i + tip_i.
   • Memoization: We use a cache (e.g., @cache in Python) to store the results of dfs(i) to avoid redundant calculations.

3. Initial Call: The maximum earnings are obtained by calling dfs(0).

Approach 2: Dynamic Programming with Binary Search

This approach is very similar to memoization but eliminates recursion by using an array to store the results.

1. Sort Rides: Sort the rides array by end point (end_i). This order ensures that when we consider a ride, we've already processed all rides that could precede it.

2. DP Array: Create a DP array f of size m + 1, where f[i] represents the maximum earnings up to the i-th ride (inclusive). f[0] is initialized to 0.

3. Iteration: Iterate through the rides. For each ride i:

   • f[i] is the maximum of:
     • f[i - 1] (skip the current ride).
     • f[j] + end_i - start_i + tip_i (take the current ride, where j is the index of the last ride whose end_j is less than or equal to start_i. This is found using binary search).

4. Result: f[m] contains the maximum earnings from all rides.

Time and Space Complexity

Both approaches have the same complexity:

• Time Complexity: O(m log m) due to the sorting step and the m binary searches (each taking O(log m) time).
• Space Complexity: O(m) for the DP array or memoization cache.

Code Examples (Python)

Approach 1 (Memoization):

from functools import cache
import bisect
 
class Solution:
    def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
        rides.sort()  # Sort by start point
 
        @cache
        def dfs(i: int) -> int:
            if i >= len(rides):
                return 0
            st, ed, tip = rides[i]
            j = bisect_left(rides, ed, lo=i + 1, key=lambda x: x[0])
            return max(dfs(i + 1), dfs(j) + ed - st + tip)
 
        return dfs(0)

Approach 2 (Dynamic Programming):

import bisect
 
class Solution:
    def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
        rides.sort(key=lambda x: x[1])  # Sort by end point
        m = len(rides)
        f = [0] * (m + 1)
        for i, (st, ed, tip) in enumerate(rides, 1):
            j = bisect_left(rides, st + 1, hi=i, key=lambda x: x[1])  # Binary search for previous ride
            f[i] = max(f[i - 1], f[j] + ed - st + tip)
        return f[m]

These examples demonstrate both approaches. The other languages (Java, C++, Go, TypeScript) would follow similar logic, adapting the specific syntax and data structures of each language. Remember that the binary search implementation might slightly vary depending on the language's standard library.